Question 1074917
R = S(Et^2/p)^1/5 
Solve for E
R/S=(Et^2)^(1/5)/p^(1/5)
(R/S)^5=Et^2/p
Et^2=p(R/S)^5
E=p(R/S)^5/t^2
----------------------------
E=(140/1.041)^5/0.025^2
=7.039 x 10^ 13
Divide by 4.184 x 10^12.
E=16.82 kilotons.
The heat capacity of the atmosphere might have been different.
The density of the air in New Mexico is likely less than 1, since the state is significantly higher than sea level. The heat capacity is closer to 1.006, which would give a value of almost exactly 20 kilotons.  He could have made a mistake in estimating the blast radius, and a small error in time would appear to be easy to make at the level of a thousandth of a second.